Default parameter in c++

Question

4.00/5 (2 votes)

See more:

Hello everyone. I got a problem with my source codes in c++. This is a problem

I got a class 'Vector2D'

class Vector2D
{
  public:
      double x;    double y;
      Vector2D():x(0.0), y(0.0)
      {}
      
      Vector(double val1, double val2):x(val1), y(val2)
      {}
}

When in another class i want to use default parameter with 'Vector2D' like this

C++

bool CanShoot(Vector2D& Target=Vector2D())
{......}

But i have an error: default argument for parameter of type 'Vector2D&' has type 'Vector2D'

How can i pass a reference parameter with a default parameter?

Posted 16-Sep-11 23:53pm

nguyenle.it

Updated 16-Sep-11 23:54pm

v2

Add a Solution

4 solutions

Solution 2

Cannot see any problem:

C++

#pragma once
#include <stdio.h>
#include <tchar.h>

class Vector2D
{
public:
  double x;    double y;
  Vector2D(double val1=0.0, double val2=0.0):x(val1), y(val2)
  {}
};

bool CanShoot(Vector2D& Target=Vector2D())
{
  printf("x: %lf, y: %lf\r\n",Target.x,Target.y);
  return 1;
}

int _tmain(int argc, _TCHAR* argv[])
{
  CanShoot();
  _gettch();
  return 0;
}

Regards.

Posted 17-Sep-11 2:35am

mbue

Comments

Philippe Mori 17-Sep-11 10:16am

Code is illegal and should not compile. See my solution.

mbue 17-Sep-11 11:39am

This isnt illegal its the same behaviour like yours but inline.
The Vector2D() isnt a ctor rather than a constructor on the stack. The compiler produces the same result like your example and the two different function calls too.
Regards.

Emilio Garavaglia 18-Sep-11 12:41pm

Philippe is right: according to the standard, you cannot bind a temporary to a reference, but only to a const-reference.
It is a well-known "Microsoft specific behavior", that non-microsoft compiler developer simply call "bug" (working differently than specified).

mbue 18-Sep-11 13:25pm

Should this be a joke? Everything on earth is temporarily. This sounds like "you can never use references except const"? Try to understand the temporary class on stack is fully accessible like other declarations, thats there is no reason to make a call to a const object. The objects lifetime is enshured during the function call.
If other compilers dont accept this construct - explain why (no links accepted)!
Regards.

Stefan_Lang 19-Sep-11 6:12am

You don't understand. In C/C++ 'temporary' is a term reserved for variables that are not explicitely listed by name in your code. instead they are created implicitely (by your compiler) as needed. Their lifetime does not exceed the statement they're used in, so if you tried to initialize a reference with one, this reference would be immediately invalidated again.

mbue 19-Sep-11 9:24am

"Their lifetime does not exceed the statement they're used in"
No you fail: the destructor of the temporary object is called AFTER the function leaves. The compiler generates two variants:
void fn(Vector2D& x){}
and
void fn(){ Vector2D x; fn(x); }
you need not to declare two different function. Accept it or leave it. Compile and debug it you will see the result is exact what you want.
Once again:
thats the code:
<pre style="margin-top: 0pt;" lang="c++">#pragma once
#include <stdio.h>
#include <tchar.h>

class Vector2D
{
public:
 double x; double y;
 Vector2D(double val1=0.0, double val2=0.0):x(val1), y(val2)
 {
 printf("constructor\r\n");
 }
 ~Vector2D()
 {
 printf("destructor\r\n");
 }
};

bool CanShoot(Vector2D& Target=Vector2D())
{
 printf("x: %lf, y: %lf\r\n",Target.x,Target.y);
 return 1;
}

int _tmain(int argc, _TCHAR* argv[])
{
 CanShoot();
 _gettch();
 return 0;
}

</pre>
thats the output:


constructor
x: 0.000000, y: 0.000000
destructor

Regards.

Stefan_Lang 19-Sep-11 11:13am

Interesting. After reading the discussion of this solution I wondered about it myself, and in two of the first hits I've found I've read exactly the answer which I transcribed here.

After your protest I checked the true source though - Quote (Bjarne Stroustrup, 'The C++ Programming Language', Third editiion, page 254):

"Unless bound to a reference or used to initialize a named object, a temporary object is destroyed at the end of the full expression in which it was created."

Seems like both references I found before omitted the first part. Doh!

My apologies.

mbue 19-Sep-11 12:25pm

it seems to be a consens ;)

Stefan_Lang 20-Sep-11 4:16am

FYI, since I still didn't fully grasp the correct use of references I looked up various articles, and it seems while indeed the temporary in this case will survive, the reference it is being assigned to should have been const. Only VC allows it to be non-const, and that is wrong by the standard. This article points out the relvant sections in the C++ (98 and 03) standard definition(s) (ignore the later part about C++0x):
http://blogs.msdn.com/b/vcblog/archive/2009/02/03/rvalue-references-c-0x-features-in-vc10-part-2.aspx

Philippe Mori 19-Sep-11 9:05am

Well, I think rule are generalized by the standard...

At least Microsoft do have the error C2664 that is used when a temporary object would be created. Thus the compiler differentiate between user temporary object and compiler generated object.

Solution 1

In your example, Vector2D() is the constructor of the class Vector2D. You cannot call the constructor explicitly. Instead, the constructor is called automatically when you instantiate an object of that class. For example:

Vector2D *foo = new Vector2D();

or with the declaration

Vector2D bar;

A class constructor has some hidden parameters that the compiler provides and you cannot.

Posted 17-Sep-11 2:24am

Chuck O'Toole

Comments

nguyenle.it 17-Sep-11 12:01pm

Thanks all, but i want to add reference parameters to the function with the default value. How can i do with my occasion above?

Emilio Garavaglia 18-Sep-11 12:49pm

This answer is misleading and claims a false concept.
Calling a constructor just creates a temporary object.
The problem is that a temporary object cannot be give to a non-const reference.

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Philippe Mori · Accepted Answer · 2011-09-17T04:20:00

According to standard, it is illegal to bind a non const reference to a temporary object. Visual Studio incorrectly allows it but give a warning at level 4.

Thus the function should be declared as:

C++

bool CanShoot(const Vector2D& Target=Vector2D()) { return true; }

However, it would be preferable to add an overload in this case:

C++

bool CanShoot()
{
    Vector2D temp;
    return CanShoot(temp);
}

This has the advantage that the "default value" can be hidden in the implementation. And it can allows some optimisations to be done.

Stefan_Lang · Accepted Answer · 2011-09-19T00:23:00

Solution 4

I like Phillipe Moris solution better, but here is an alternate:

C++

class Vector2D {
   // ... your stuff goes here
};
const Vector2D null_vector; // implicitely calls default constructor

bool CanShoot(const Vector2D& target = null_vector) {
   // ... your code here
}

Posted 19-Sep-11 0:23am

Stefan_Lang

Comments

Philippe Mori 19-Sep-11 9:19am

I do like that solution. The variable could also be made static. That solution would be better when there are multiple defaults depending on the function.

nguyenle.it 19-Sep-11 18:57pm

It work like a charm, thank you so much. and thank you everyone for giving me the solution.

Default parameter in c++

4 solutions

Solution 3

Solution 4

Solution 2

Solution 1

Add your solution here

Preview 0

Existing Members

...or Join us